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A271591 Second most significant bit of the tribonacci number A000073(n).

%I A271591 #34 Feb 07 2018 18:14:46
%S A271591 0,0,1,1,1,0,0,0,0,1,1,1,1,0,0,0,0,0,1,1,1,0,0,0,0,0,1,1,1,0,0,0,0,0,
%T A271591 1,1,1,1,0,0,0,0,1,1,1,1,0,0,0,0,0,1,1,1,0,0,0,0,0,1,1,1,0,0,0,0,0,1,
%U A271591 1,1,1,0,0,0,0,0,1,1,1,0,0,0,0,0,1,1,1,0,0,0,0,0,1,1,1,0,0,0,0,0,1,1,1,1,0,0,0,0,0,1,1
%N A271591 Second most significant bit of the tribonacci number A000073(n).
%C A271591 It is conjectured that after the first two 0's, the number of consecutive 0's is only 4 or 5, and the number of consecutive 1's is only 3 or 4 (tested up to n=10^4). The sequence looks quasiperiodic (or with a very long true period if any).
%H A271591 Chai Wah Wu, <a href="/A271591/b271591.txt">Table of n, a(n) for n = 4..10000</a>
%F A271591 a(n) = floor(A000073(n)/(2^(ceiling(log_2(A000073(n) + 1)) - 2))) - 2.
%F A271591 a(n) = A079944(A000073(n)-2). - _Michel Marcus_, Apr 22 2016
%e A271591 (Second MSB in parenthesis)
%e A271591   n   A000073(n)      A000073(n)
%e A271591       decimal         binary
%e A271591   4      2        ->   1(0)
%e A271591   5      4        ->   1(0)0
%e A271591   6      7        ->   1(1)1
%e A271591   7      13       ->   1(1)01
%e A271591   8      24       ->   1(1)000
%e A271591   9      44       ->   1(0)1100
%e A271591   10     81       ->   1(0)10001
%e A271591   11     149      ->   1(0)010101
%t A271591 a = LinearRecurrence[{1, 1, 1}, {0, 0, 1}, 120];(* to generate A000073 *)
%t A271591 Table[IntegerDigits[a, 2][[i]][[2]], {i, 5, Length[a]}]
%o A271591 (Python)
%o A271591 A271591_list, a, b, c = [], 0, 1 ,1
%o A271591 for n in range(4,10001):
%o A271591     a, b, c = b, c, a+b+c
%o A271591     A271591_list.append(int(bin(c)[3])) # _Chai Wah Wu_, Feb 07 2018
%Y A271591 Cf. A000073 (tribonacci numbers), A079944 (2nd msb), A272170.
%K A271591 nonn,base
%O A271591 4,1
%A A271591 _Andres Cicuttin_, Apr 10 2016